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Sagot :
Answer:
a) E(x) = -2p^2 + 2p + 2
b) Number is maximized when p = 1/2
Step-by-step explanation:
Determine the Expected number of games when ( i ) = 2
The number of possible combinations that both teams win two games :
AA, BB, ABB, ABA, BAA, BAB = 6 combinations
P( team A winning ) = p
P( team B wins ) = 1 - p
Attached below is the detailed solution on the expected number of games
expected number of games ; E(x) = -2p^2 + 2p + 2
ii) Number is maximized when p = 1/2
In this exercise we will use the knowledge of probability and combination, so we have what will be:
a)[tex]E(x) = -2p^2 + 2p + 2[/tex]
b)[tex]p = 1/2[/tex]
Organizing the information given in the statement as:
- Expected number of games when ( i ) = 2
A)The number of possible combinations that both teams win two games :
[tex]AA, BB, ABB, ABA, BAA, BAB = 6 \ combinations\\P( team\ A \ winning ) = p\\P( team \ B \ wins ) = 1 - p\\E(x) = -2p^2 + 2p + 2[/tex]
B) To calculate the maximum number we must solve the quadratic equation, like this:
[tex]p=1/2[/tex]
See more about probability at brainly.com/question/795909
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